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<h1 class="heading"><a href="MATH-2023-OPDE.html"><span class="title">MATH 2023: Ordinary and Partial Differential Equations</span></a></h1>
<p class="byline">Xiaoyi Chen and Wei Zhang</p>
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<a href="ch_first.html" data-scroll="ch_first" class="internal"><span class="codenumber">1</span> <span class="title">Introduction</span></a><ul>
<li><a href="sec_1-intro.html" data-scroll="sec_1-intro" class="internal">Classification of Differential Equations</a></li>
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<a href="ch_second.html" data-scroll="ch_second" class="internal"><span class="codenumber">2</span> <span class="title">First Order Ordinary Differential Equations</span></a><ul>
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<li><a href="sec2_2.html" data-scroll="sec2_2" class="internal">Further Discussion of Linear Equations (For reading only)</a></li>
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<li><a href="sec3_1.html" data-scroll="sec3_1" class="internal">Homogeneous equations with constant coefficient</a></li>
<li><a href="sec3_2.html" data-scroll="sec3_2" class="internal">Fundamental Solutions of Linear Homogeneous Equations</a></li>
<li><a href="sec3_3.html" data-scroll="sec3_3" class="internal">Linear Independence and Wronskian</a></li>
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<a href="ch_five.html" data-scroll="ch_five" class="internal"><span class="codenumber">5</span> <span class="title">Series Solutions of Second Order Linear Equations</span></a><ul>
<li><a href="sec5_1.html" data-scroll="sec5_1" class="internal">Brief Review on Power Series</a></li>
<li><a href="sec5_2.html" data-scroll="sec5_2" class="internal">Introduction</a></li>
<li><a href="sec5_3.html" data-scroll="sec5_3" class="internal">Series Solutions Near an Ordinary Point</a></li>
<li><a href="sec5_4.html" data-scroll="sec5_4" class="internal">Euler’s Equation</a></li>
<li><a href="sec5_5.html" data-scroll="sec5_5" class="active">Series Solution near a Regular Singular Point</a></li>
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<a href="ch_six.html" data-scroll="ch_six" class="internal"><span class="codenumber">6</span> <span class="title">System of First Order Linear Equations</span></a><ul>
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<a href="ch_seven.html" data-scroll="ch_seven" class="internal"><span class="codenumber">7</span> <span class="title">Partial Differential Equations</span></a><ul>
<li><a href="sec7_1.html" data-scroll="sec7_1" class="internal">Two-Point Boundary Value Problems</a></li>
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<li><a href="sec7_5.html" data-scroll="sec7_5" class="internal">Even and Odd Functions</a></li>
<li><a href="sec7_6.html" data-scroll="sec7_6" class="internal">Introduction to Partial Differential Equations</a></li>
<li><a href="sec7_7.html" data-scroll="sec7_7" class="internal">1D Heat Equation; Solutions by Separation of Variable and Fourier Series</a></li>
<li><a href="sec7_8.html" data-scroll="sec7_8" class="internal">Other Heat Conduction Problems</a></li>
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<li><a href="sec8_1.html" data-scroll="sec8_1" class="internal">What are Laplace Transforms, and Why?</a></li>
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<li><a href="sec8_5.html" data-scroll="sec8_5" class="internal">Step input and Impulse problems</a></li>
<li><a href="sec8_6.html" data-scroll="sec8_6" class="internal">Laplace transform for PDE (heat equation)</a></li>
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<li class="link"><a href="solutions-1.html" data-scroll="solutions-1" class="internal"><span class="codenumber">A</span> <span class="title">Selected Hints</span></a></li>
<li class="link"><a href="solutions-2.html" data-scroll="solutions-2" class="internal"><span class="codenumber">B</span> <span class="title">Selected Solutions</span></a></li>
<li class="link"><a href="appendix-1.html" data-scroll="appendix-1" class="internal"><span class="codenumber">C</span> <span class="title">List of Symbols</span></a></li>
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<main class="main"><div id="content" class="pretext-content"><section class="section" id="sec5_5"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">5.5</span> <span class="title">Series Solution near a Regular Singular Point</span>
</h2>
<p id="p-230">Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq5_11">
\begin{equation}
P(x) y^{\prime \prime}+Q(x) y^{\prime}+R(x) y=0,\quad x&gt;0.\tag{5.5.1}
\end{equation}
</div>
<p class="continuation">Suppose that <span class="process-math">\(x=0\)</span> is a regular singular point. (Note, if the point is <span class="process-math">\(x=x_0\text{,}\)</span> let <span class="process-math">\(t=x-x_0\text{,}\)</span> then the corresponding point is changed to <span class="process-math">\(t=0\text{.}\)</span> ) Then <span class="process-math">\(x Q(x)/P(x)\)</span> and <span class="process-math">\(x^2 R(x)/P(x)\)</span> are analytic at <span class="process-math">\(x=0\text{.}\)</span> Suppose that</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq5_12">
\begin{equation}
p(x)=x \frac{Q(x)}{P(x)}=\sum_{n=0}^{\infty} p_n x^n,\quad q(x)=x^2 \frac{R(x)}{P(x)}=\sum_{n=0}^{\infty} q_n x^n,\quad 0&lt;x&lt;\rho,\tag{5.5.2}
\end{equation}
</div>
<p class="continuation">where</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
p_0=\lim \limits_{x \to 0} x \frac{Q(x)}{P(x)},\quad q_0=\lim \limits_{x \to 0} x^2 \frac{R(x)}{P(x)}.
\end{equation*}
</div>
<p class="continuation">Seek a solution in the form</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq5_13">
\begin{equation}
y=\sum_{n=0}^{\infty} a_n x^{n+r}=a_0 x^r+a_1 x^{1+r}+a_2 x^{2+r}+\cdots,\tag{5.5.3}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(a_0 \neq 0\text{.}\)</span></p>
<p id="p-231">Write (<a href="" class="xref" data-knowl="./knowl/eq5_11.html" title="Equation 5.5.1">(5.5.1)</a>) as</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_11.html ./knowl/eq5_12.html ./knowl/eq5_13.html ./knowl/eq5_14.html ./knowl/eq5_15.html ./knowl/eq5_17.html" id="eq5_14">
\begin{equation}
x^2 y^{\prime \prime}+ x \left( x \frac{Q(x)}{P(x)}\right)y^{\prime}+\left(x^2 \frac{R(x)}{P(x)} \right) y=0.\tag{5.5.4}
\end{equation}
</div>
<p class="continuation">Substituting (<a href="" class="xref" data-knowl="./knowl/eq5_12.html" title="Equation 5.5.2">(5.5.2)</a>) and (<a href="" class="xref" data-knowl="./knowl/eq5_13.html" title="Equation 5.5.3">(5.5.3)</a>) into the (<a href="" class="xref" data-knowl="./knowl/eq5_14.html" title="Equation 5.5.4">(5.5.4)</a>):</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_11.html ./knowl/eq5_12.html ./knowl/eq5_13.html ./knowl/eq5_14.html ./knowl/eq5_15.html ./knowl/eq5_17.html">
\begin{equation*}
\begin{aligned}
&amp;x^2 \left[r(r-1) a_0 x^{r-2}+r (r+1) a_1 x^{r-1}+\cdots \right]+x (p_0+p_1 x+\cdots) \left[r a_0 x^{r-1}+(r+1) a_1 x^r+\cdots \right]\\
&amp;\quad +(q_0+q_1 x+\cdots)\left[ a_0 x^r+a_1 x^{1+r}+\cdots \right]=0\\
&amp; \to r(r-1) a_0 x^r+(r+1) r a_1 x^{r+1}+\cdots\\
&amp;\qquad +r p_0 a_0 x^r+[r p_1 a_0+(r+1) p_0 a_1]x^{r+1}+\cdots\\
&amp;\qquad +q_0 a_0 x^r+(q_1 a_0+q_0 a_1) x^{r+1}+\cdots=0
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Dividing both sides by the factor <span class="process-math">\(x^r\text{:}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_11.html ./knowl/eq5_12.html ./knowl/eq5_13.html ./knowl/eq5_14.html ./knowl/eq5_15.html ./knowl/eq5_17.html">
\begin{equation*}
a_0+[(r+1) r a_1+r p_1 a_0+(r+1) p_0 a_1+q_1 a_0+q_0 a_1] x+\cdots=0,
\end{equation*}
</div>
<p class="continuation">which further gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_11.html ./knowl/eq5_12.html ./knowl/eq5_13.html ./knowl/eq5_14.html ./knowl/eq5_15.html ./knowl/eq5_17.html" id="eq5_15">
\begin{equation}
[r^2+r( p_0-1)+q_0] a_0=0,\tag{5.5.5}
\end{equation}
</div>
<p class="continuation">and</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_11.html ./knowl/eq5_12.html ./knowl/eq5_13.html ./knowl/eq5_14.html ./knowl/eq5_15.html ./knowl/eq5_17.html" id="eq5_16">
\begin{equation}
(r+1) r a_1+r p_1 a_0+(r+1) p_0 a_1+q_1 a_0+q_0 a_1=0.\tag{5.5.6}
\end{equation}
</div>
<p class="continuation">From (<a href="" class="xref" data-knowl="./knowl/eq5_15.html" title="Equation 5.5.5">(5.5.5)</a>), since <span class="process-math">\(a_0 \neq 0\text{,}\)</span> we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_11.html ./knowl/eq5_12.html ./knowl/eq5_13.html ./knowl/eq5_14.html ./knowl/eq5_15.html ./knowl/eq5_17.html" id="eq5_17">
\begin{equation}
r^2+r( p_0-1)+q_0=0.\tag{5.5.7}
\end{equation}
</div>
<p class="continuation">Equation (<a href="" class="xref" data-knowl="./knowl/eq5_17.html" title="Equation 5.5.7">(5.5.7)</a>) is the <dfn class="terminology">indicial equation</dfn> and the <span class="process-math">\(r\)</span> is called the <dfn class="terminology">exponents at the singularity</dfn>.</p>
<p id="p-232">Therefore, suppose (<a href="" class="xref" data-knowl="./knowl/eq5_17.html" title="Equation 5.5.7">(5.5.7)</a>) has two real roots <span class="process-math">\(r_1\)</span> and <span class="process-math">\(r_2\)</span> (<span class="process-math">\(r_1 \geq r_2\)</span>). Then, there exists a solution of the form</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_17.html" id="eq5_18">
\begin{equation}
y_1=x^{r_1} \sum_{n=0}^{\infty} a_n x^n \quad \textrm{with}~a_0=1, \quad \textrm{for}~0&lt;x&lt;\rho.\tag{5.5.8}
\end{equation}
</div>
<p id="p-233">For the second linear independent solution, there are three cases:Case 1 <span class="process-math">\(r_1-r_2 \neq \mathbb{Z}\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_18.html ./knowl/eq5_19.html ./knowl/eq5_11.html">
\begin{equation*}
y_2=x^{r_2} \sum_{n=0}^{\infty} b_n x^n \quad \textrm{with}~b_0=1 \quad \textrm{for}~0&lt;x&lt;\rho.
\end{equation*}
</div>
<p class="continuation">Case 2 <span class="process-math">\(r_1-r_2=0\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_18.html ./knowl/eq5_19.html ./knowl/eq5_11.html">
\begin{equation*}
y_2=y_1(x) \ln x+x^{r_1} \sum_{n=1}^{\infty} c_n x^n,\quad \textrm{for}~0&lt;x&lt;\rho.
\end{equation*}
</div>
<p class="continuation">Case 3 <span class="process-math">\(r_1-r_2=\mathbb{Z}\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_18.html ./knowl/eq5_19.html ./knowl/eq5_11.html" id="eq5_19">
\begin{equation}
y_2(x)=a y_1(x) \ln x+x^{r_2} \sum_{n=0}^{\infty} d_n x^n \quad \textrm{with}~d_0=1 \quad \textrm{for}~0&lt;x&lt;\rho.\tag{5.5.9}
\end{equation}
</div>
<p class="continuation">In (<a href="" class="xref" data-knowl="./knowl/eq5_18.html" title="Equation 5.5.8">(5.5.8)</a>) to (<a href="" class="xref" data-knowl="./knowl/eq5_19.html" title="Equation 5.5.9">(5.5.9)</a>), <span class="process-math">\(a_n, b_n, c_n, d_n,  (n \geq 1)\)</span> and <span class="process-math">\(a\)</span> can be determined by substituting the solution into (<a href="" class="xref" data-knowl="./knowl/eq5_11.html" title="Equation 5.5.1">(5.5.1)</a>).</p>
<p id="p-234"><dfn class="terminology">Example 1</dfn> (The Bessel Equation)Find the two linear independent solutions of the Bessel equation of order <span class="process-math">\(1/2\text{:}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq5_21">
\begin{equation}
x^2 y^{\prime \prime}+x y^{\prime}+[x^2-\underline{(1/2)}^2]y=0.\tag{5.5.10}
\end{equation}
</div>
<p class="continuation">Note <span class="process-math">\(1/2\)</span> is the order of the Bessel equation.</p>
<p id="p-235"><dfn class="terminology">Solution</dfn>:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
p(x)=x \frac{Q(x)}{P(x)}=x \frac{x}{x^2}=1,\quad q(x)=x^2 \frac{R(x)}{P(x)}=x^2 \frac{x^2-(1/2)^2}{x^2}=-\frac{1}{4}+x^2.
\end{equation*}
</div>
<p class="continuation">Since both <span class="process-math">\(x Q(x)/P(x)\)</span> and <span class="process-math">\(x^2 R(x)/P(x)\)</span> are analytic at <span class="process-math">\(x=0\text{,}\)</span> <span class="process-math">\(x=0\)</span> is a regular singular point.</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
p_0=\lim \limits_{x \to 0} x \frac{Q(x)}{P(x)}=1,\quad q_0=\lim \limits_{x \to 0} x^2 \frac{R(x)}{P(x)}=-\frac{1}{4}.
\end{equation*}
</div>
<p class="continuation">The indicial equation is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
r^2+(p_0-1) r+q_0=0 \to r^2-r+r-\frac{1}{4}=0 \to r^2-\frac{1}{4}=0 \to r_1=\frac{1}{2},\quad r_2=-\frac{1}{2}.
\end{equation*}
</div>
<p id="p-236"><span class="process-math">\(r_1-r_2=1\)</span> corresponds to Case 3 of the theorem. According to the theorem, the first solution is given by</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html" id="eq5_20">
\begin{equation}
y_1=x^{r_1} \sum_{n=0}^{\infty} a_n x^n=\sum_{n=0}^{\infty} a_n x^{n+1/2} \quad \textrm{with}~a_0=1.\tag{5.5.11}
\end{equation}
</div>
<p class="continuation">Substituting (<a href="" class="xref" data-knowl="./knowl/eq5_20.html" title="Equation 5.5.11">(5.5.11)</a>) into (<a href="" class="xref" data-knowl="./knowl/eq5_21.html" title="Equation 5.5.10">(5.5.10)</a>):</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation*}
\begin{aligned}
&amp;x^2 \sum_{n=0}^{\infty} (n+1/2) (n-1/2) a_n x^{n-3/2}+x \sum_{n=0}^{\infty}(n+1/2) a_n x^{n-1/2}+(x^2-1/4) \sum_{n=0}^{\infty} a_n x^{n+1/2}=0,\\
&amp;\to \sum_{n=0}^{\infty} (n^2-1/4) a_n x^{n+1/2}+\sum_{n=0}^{\infty} (n+1/2) a_n x^{n+1/2}+\sum_{n=0}^{\infty} a_n x^{n+5/2}-\frac{1}{4} \sum_{n=0}^{\infty} a_n x^{n+1/2}=0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Dividing both sides by the factor <span class="process-math">\(x^{1/2}\text{:}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation*}
\begin{aligned}
&amp;\to \sum_{n=0}^{\infty} (n^2-1/4) a_n x^n+\sum_{n=0}^{\infty} (n+1/2) a_n x^n+\sum_{n=0}^{\infty} a_n x^{n+2}-\frac{1}{4} \sum_{n=0}^{\infty} a_n x^n=0\\
&amp;\to \sum_{n=0}^{\infty} [n^2-\frac{1}{4}+n+\frac{1}{2}-\frac{1}{4}] a_n x^n+\sum_{n=0}^{\infty} a_n x^{n+2}=0\\
&amp;\to \sum_{n=0}^{\infty} n(n+1) a_n x^n+\sum_{k=2}^{\infty}a_{k-2} x^k=0\\
&amp;\to 0+1 \times 2 \times a_1 x+\sum_{n=2}^{\infty} n(n+1) a_n x^n+\sum_{n=2}^{\infty} a_{n-2} x^n=0\\
&amp;\to 2 a_1 x+\sum_{n=2}^{\infty} [n(n+1) a_n+a_{n-2}] x^n=0\\
&amp;\to a_1=0,\quad n(n+1)a_n+a_{n-2}=0,\\
&amp;\to  a_1=0, \quad a_n=-\frac{a_{n-2}}{n(n+1)},\quad n \geq 2.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Based on this recurrence relation, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation*}
\begin{aligned}
&amp;n=2:\quad a_2=-\frac{a_0}{3 \cdot 2}=-\frac{1}{3!},\\
&amp;n=3: \quad a_3=0,\\
&amp;n=4: \quad a_4=-\frac{a_2}{5\cdot 4}=+\frac{1}{5!},\\
&amp;n=5:\quad a_5=0,\\
&amp;n=6: \quad a_6=-\frac{a_4}{7 \cdot 6}=-\frac{1}{7!},\\
&amp;\cdots\\
&amp;a_{2k+1}=0,\\
&amp;a_{2k}=(-1)^k \frac{1}{(2k+1)!}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation*}
\begin{aligned}
y_1&amp;=\sum_{n=0}^{\infty} a_n x^{n+1/2}=\sum_{k=0}^{\infty} a_{2k} x^{2k+1/2}
+\sum_{k=0}^{\infty} a_{2k+1}x^{2k+1+1/2}\\
&amp;=\sum_{n=0}^{\infty} (-1)^k \frac{x^{2k+1/2}}{(2k+1)!}=x^{-1/2} \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}\\
&amp;=x^{-1/2} \sin x.
\end{aligned}
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="" id="p-237">
\begin{equation*}
\left( \frac{2}{\pi} \right)^{1/2} y_1=\left(  \frac{2}{\pi x}    \right)^{1/2} \sin x \xlongequal[\quad ]{\textrm{denoted as}}J_{1/2}(x),
\end{equation*}
</div>
<p class="continuation">is also a solution, which is called Bessel function of order <span class="process-math">\(1/2\text{.}\)</span></p>
<p id="p-238">According to the theorem, the second solution is given by</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=a y_1 \ln x+x^{r_2} \sum_{n=0}^{\infty} d_n x^n=a y_1 \ln x+\sum_{n=0}^{\infty} d_n x^{n-1/2} \quad \textrm{with}~d_0=1.
\end{equation*}
</div>
<p class="continuation">Substituting this solution into the ODE, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;a \ln x [\underline{x^2 y_1^{\prime \prime}+x y_1^{\prime}+(x^2-1/4) y_1}]+2 a x y_1^{\prime}+\sum_{n=0}^{\infty} (n^2-n) d_n x^{n-1/2}+\sum_{n=0}^{\infty} d_n x^{n+3/2}=0,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where the term with underline is zero. We also have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
2 a x y_1^{\prime}&amp;=2 a x \left( \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1/2}}{(2n+1)!}   \right)^{\prime}\\
&amp;=2 a x \cdot \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1/2)  x^{2n-1/2}}{(2n+1)!}\\
&amp;=2 a \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1/2)  x^{2n+1/2}}{(2n+1)!}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Therefore,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
2 a \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1/2)  x^{2n+1/2}}{(2n+1)!}+\sum_{n=0}^{\infty} (n^2-n) d_n x^{n-1/2}+\sum_{k=2}^{\infty} d_{k-2} x^{k-1/2}=0.
\end{equation*}
</div>
<p class="continuation">Multiplying both sides by the factor <span class="process-math">\(x^{1/2}\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq5_22">
\begin{equation}
\begin{aligned}
&amp;2 a \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1)  x^{2n+1}}{(2n+1)!}+\sum_{n=0}^{\infty} (n^2-n) d_n x^n+\sum_{k=2}^{\infty} d_{k-2} x^k=0\\
&amp;\to 2 a \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1)  x^{2n+1}}{(2n+1)!}+\sum_{n=2}^{\infty} [n(n-1)d_n+d_{n-2}] x^n=0.
\end{aligned}\tag{5.5.12}
\end{equation}
</div>
<p class="continuation">Consider the first few terms:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;2 a \frac{1}{1!} x+2 a(-1) \frac{1}{2!} x^3+\cdots+[(2 \cdot 1 d_2+d_0) x^2+(3 \cdot 2 d_3+d_1) x^3+\cdots]=0\\
&amp;\to 2 a x+(2 d_2+d_0) x^2+\left[-a+(6d_3+d_1)\right]x^3+\cdots=0,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">which gives</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
a=0,\quad 2d_2+d_0=0,\cdots.
\end{equation*}
</div>
<p id="p-240">Then <span class="process-math">\((\ref{eq5_22})_2\)</span> becomes</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\sum_{n=0}^{\infty}[n (n-1)d_n+d_{n-2}] x^n=0,
\end{equation*}
</div>
<p class="continuation">which gives the recurrence relation</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;d_n=-\frac{d_{n-2}}{n(n-1)},\quad n\geq 2\\
&amp; d_{2k}=(-1)^k \frac{1}{(2 k)!},\quad d_{2k+1}=(-1)^k \frac{d_1}{(2k+1)!},
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(d_1\)</span> is arbitrary. We set <span class="process-math">\(d_1=0\)</span> so that <span class="process-math">\(d_{2k+1}=0\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
y_2 &amp;=a y_1 \ln x+\sum_{n=0}^{\infty} d_n x^{n-1/2}\\
&amp;=0+x^{-1/2} \left[ \sum_{k=0}^{\infty} d_{2k} x^{2k}+\sum_{k=0}^{\infty} d_{2k+1} x^{2k+1}   \right]\\
&amp;=x^{-1/2} \sum_{k=0}^{\infty} (-1)^k \frac{x^{2 k}}{(2 k)!}=x^{-1/2} \cos x.
\end{aligned}
\end{equation*}
</div>
<p id="p-241"><span class="process-math">\(\left(\frac{2}{\pi} \right)^{1/2} y_1=\left(\frac{2}{\pi x}\right)^{1/2} \cos x \xlongequal[ ]{\textrm{denoted as}} J_{-1/2}(x)\)</span> is also a solution, which is called the Bessel function of order <span class="process-math">\(-1/2\text{.}\)</span> Thus, the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=C_1 J_{1/2}(x)+C_2 J_{-1/2}(x).
\end{equation*}
</div>
<p id="p-242"><dfn class="terminology">Note</dfn>(i) For a regular singular point, the corresponding power series solutions are called Frobenius solutions.(ii) The procedures to obtain series solution:(1) Identify the point concerned is an ordinary point or regular singular point.(2) Ordinary point: The independent solutions are given by</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=\sum_{n=0}^{\infty} a_n (x-x_0)^n=a_0 y_1+a_1 y_2 .
\end{equation*}
</div>
<p class="continuation">For <span class="process-math">\(y_1\text{,}\)</span> setting <span class="process-math">\(a_0=1, a_1=0\text{.}\)</span>For <span class="process-math">\(y_2\text{,}\)</span> setting <span class="process-math">\(a_0=0, a_1=1\text{.}\)</span>(3) Regular singular point at <span class="process-math">\(x_0\text{:}\)</span>A change of variables: <span class="process-math">\(t=x-x_0\text{,}\)</span> which changes the regular singular point to <span class="process-math">\(t=0\text{.}\)</span>(3 a) Find <span class="process-math">\(p_0\)</span> and <span class="process-math">\(q_0\)</span> and solve the indicial equations:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
r^2 +r (p_0-1)+q_0=0
\end{equation*}
</div>
<p class="continuation">to obtain two roots <span class="process-math">\(r_1\)</span> and <span class="process-math">\(r_2\)</span> (set <span class="process-math">\(r_1 \geq r_2\)</span>).(3 b) Construct the first solution as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=t^{r_1} \sum_{n=0}^{\infty} a_n t^n
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(a_0=1\text{.}\)</span>(3 c) To construct the second solution: there are three cases.</p></section></div></main>
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